I do not know what the question is but this is my favorite thing to use when working with these clocks problems.

If the hands art both upon 12 then,
The minute hand moves 6 degrees per minute.
The hour hand moves .5 degrees per minute.
Thus the difference is 5.5 degree per minute.
Thus, after t=>0 minutes the distance between them is 5.5t (of course if they do not travel all the way around).

If you do consider the difficulty when they travel all the way around then the difference is (5.5t)_{360} that means the remainder that is left after subtracting 360 sufficiently enough times.
For example (750)_{360}=30 because we subtract 360 twice and have a number less then 360 which we need.

The hour-hand of the clock moves around the clock in 12 hours. During this time it is overtaken by the minute-hand 11 times.
That means (1 + 1/11) hour after the last meeting the hands will meet again.